Exercise Set 1 Answers
These exercises cover the topics of Algebra, Equations and Inequalities.
Simplify each of the following (Hint: use the rules of exponents where needed).
- \(x = 3pq+5pr=2qr+qp-6rp\)
- \(y = 5l^2mn+2nl^2m-3mln^2+l^2nm+4n^2ml-nm^2\)
- \(z = \dfrac{(s^\frac{1}{3})^\frac{3}{4}\times (t^\frac{1}{4})^{-1}}{(t^\frac{1}{2}\times (s^{-\frac{1}{4}})^{-1})}\)
Answers:
We collect the so-called “like terms” (if there are any): \[\begin{align*} x &= 3pq + 5pr - 2qr + qp -6rp = 3pq +qp +5pr - 6rp - 2qr \\ &= 4pq - pr - 2qr \quad\text{ we add and/or subtract ``like terms''} \end{align*}\]
We collect the so-called “like terms” (if there are any): \[\begin{align} y &= 5l^2 mn + 2nl^2m - 3mln^2 + l^2nm + 4n^2ml - nm^2\\ &= 5l^2 mn + 2nl^2m + l^2nm - 3mln^2 + 4n^2ml - nm^2\\ &= 8l^2 mn + n^2lm - nm^2 \quad\text{ we add and/or subtract ``like terms''} \end{align}\]
We first complete the calculation in both the numerator and the denominator using the laws & rules of indices: \[\begin{align} z &= \frac{(s^\frac{1}{3})^\frac{3}{4} \times (t^\frac{1}{4})^{-1}}{(t^\frac{1}{2})^4 \times (s^{-\frac{1}{4}})^{-1}}=\frac{s^{\frac{1}{3}\times\frac{3}{4}}\times t^{-\frac{1}{4}}}{t^{\frac{1}{2} \times 4}\times s^{\frac{1}{4}}}\\ &= \frac{s^\frac{1}{4}\times t^{-\frac{1}{4}}}{t^2 \times s^\frac{1}{4}}\quad \text{we treat the powers of the same base together using laws of indices}\\ &= s^{\frac{1}{4}-\frac{1}{4}} \times t^{-\frac{1}{4}-2} = 1 \times t^{-\frac{9}{4}} = \frac{1}{t^\frac{9}{4}} \end{align}\]
It is best to write your final answer with positive exponents only.
Expand the brackets in each of the following and simplify the expression.
- \(-4x(2x-y)(3x+2y)\)
- \((a-2b)(2a-3b)(3a-4b)\)
- \(-\{-2[x-3(y-4)]-5(z+6)\}\)
- \((v^3-v^2-2)(1-3v+2v^2)\)
Answers:
We have three factors. First we multiply \(-4x\) by \(2x-y\), then we multiply the outcome by \((3x + 2y)\). \[\begin{align*} -4x(2x-y)(3x+2y) &= (-8x^2 + 4xy)(3x+2y)\\ & = -24x^3 - 4x^2y + 8xy^2. \end{align*}\]
Again, we have three factors, we repeat the same process (make sure you simplify the outcome in each step, i.e. add/subtract like-terms): \[\begin{align*} (a-2b)(2a-3b)(3a-4b) &= (2a^2 - 3ab - 4ab + 6b^2)(3a - 4b)\\ &= (2a^2 - 7ab + 6b^2)(3a - 4b)\\ &= 6a^3 - 8a^2 b - 21a^2b + 28ab^2 + 18b^2a - 24b^3\\ &= 6a^3 - 29a^2b + 46b^2a - 24b^3 \end{align*}\]
We calculate the inner brackets first, then we remove the brackets. Lastly, we add/subtract like terms: \[\begin{align*} -\{ -2[x - 3(y-4)]- 5(z+6)\} &= - \{-2[x - 3y + 12] - 5(z+6)\}\\ &= - (-2x + 6y - 24 -5z -30)\\ &= -(-2x + 6y -5z -54)\\ &= 2x - 6y + 5z +54 \end{align*}\]
\[\begin{align*} (v^3 - v^2 - 2)(1 - 3v + 2v^2) &= v^3 - 3v^4 + 2v^5 - v^2 + 3v^3 - 2v^4 - 2 + 6v - 4v^2\\ &= 2v^5 - 5v^4 + 4v^3 - 5v^2 + 6v - 2 \end{align*}\]
Simplify each of the following.
- \(\dfrac{p}{q^3}\div\dfrac{p^3}{q}\)
- \(\dfrac{a^2b}{2c}\times\dfrac{ac^2}{2b}\div\dfrac{b^2c}{2a}\)
- \(\dfrac{8x^{-3}\times 3x^2}{6x^{-4}}\)
- \(\dfrac{3x}{3x^2+6x}\)
Answers:
\[ \frac{p}{q^3} \times \frac{q}{p^3} = \frac{pq}{q^3 p^3} = \frac{1}{p^2 q^2} \]
We first multiply the first two fractions, then we divide the outcome by the third \[\begin{align*} \left(\frac{a^2 b}{2c} \times \frac{ac^2}{2b}\right)\div \frac{b^2c}{2a} &= \frac{a^3 b c^2}{4cb}\div \frac{b^2 c}{2a}\\ &= \frac{a^3 b c^2}{4cb}\times \frac{2a}{b^2c}\\ &= \frac{2a^4 b c^2}{4b^3 c^2} = \frac{a^4}{2b^2} \end{align*}\]
\[\begin{align*} \frac{8x^{-3} \times 3x^2}{6x^{-4}} &= \frac{24x^{-3+2}}{6x^{-4}}\\ &= 4x^{-1+4}\\ &= 4x^3 \end{align*}\]
\[ \frac{3x}{3x^2 + 6x} = \frac{3x}{3x(x+2)} = \frac{1}{x+2} \]
Factorise the following expressions.
- \(18x^2y-12xy^2\)
- \(x^3+4x^2y-3xy^2-12y^3\)
- \(25x^2-4y^2\)
Answers:
We look for the common factors between the two terms, in this case \(6xy\) \[ 18x^2y - 12xy^2 = 6xy(3x-2y) \]
This expression has 4 terms, we divide them into 2 groups each of 2 terms, then factorise these groups and use the difference of two squares to factorise the term with squares \[\begin{align*} x^3 + 4x^2y - 3xy^2 - 12y^3 &= x^2(x+4y) - 3y^2(x+4y)\\ &= (x+4y)(x^2-3y^2)\\ &= (x+4y)(x-\sqrt{3}y)(x+\sqrt{3}y) \end{align*}\]
This is another application of the difference between two squares identity \(a^2 - b^2 = (a-b)(a+b)\) \[ 25x^2 - 4y^2 = (5x+2y)(5x-2y) \]
The characteristic equation of a perfect gas is given by \(PV = mRT\) where \(m\) is the mass, \(P\) is the pressure, \(V\) is the volume, \(T\) is the temperature and \(R\) is the universal gas constant. Make temperature the subject of the formula.
Answer:
\[PV = mRT\] Divide both sides by \(mR\), then \[T = \frac{PV}{mR}\]
The airflow over a turbine blade causes drag \(D\), which is given by \(D= \frac{\rho C v^2 A}{2}\), where \(\rho\) is fluid density, \(C\) is the drag coefficient, \(v\) is fluid velocity and \(A\) is the frontal area of the blade. Make the frontal area the subject of the formula.
Answer:
First multiply both sides by 2, then divide both sides of the outcome by \(\rho C v^2\)
\[\begin{align*} D &= \frac{\rho C v^2 A}{2}\\ 2D &= \rho C v^2 A\\ A &= \frac{2D}{\rho C v^2} \end{align*}\]
Make \(b\) the subject of the following formula. \[W=\frac{t\sqrt{a+b^2}}{2\pi}\]
Answer:
First, multiply both sides by \(2\pi\) and divide both sides by \(t\), then square both sides and make \(b^2\) the subject
\[\begin{align*} W &= \frac{t\sqrt{a + b^2}}{2\pi}\\ \frac{2\pi}{t} &= \sqrt{a+b^2}\\ \left(\frac{2\pi}{t}\right)^2 &= a + b^2\\ b^2 &= \frac{4\pi^2}{t^2} - a \end{align*}\]
Now it is necessary to take the square root of both sides, recalling that there is both a positive and a negative root: \[ b = \pm \sqrt{\frac{4\pi^2}{t^2} - a} \] Often in applications such a quantity may have a physical significance that will force it to be only the positive root or only the negative root, for example if the quantity \(b\) represented a length then we would take the positive solution.